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zepelin [54]
3 years ago
14

What happens to a liquid when it loses energy

Physics
1 answer:
NeTakaya3 years ago
8 0
It can solidify, it depends on the tempeture.
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The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear
WITCHER [35]

Answer:

3234.2 W

Explanation:

Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.

So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².

Now, the power radiated on the patch of area 0.570 m² at the equator is

P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W

6 0
2 years ago
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive
olganol [36]

Answer:

The force is     F_c  =  789.03 \  N    

Explanation:

From the question we are told that

   The tangential  resistive force is F_t  =   115 \ N

   The mass of the wheel is  m  = 1.80 kg

  The diameter of the wheel is  d =  50.0 cm  = 0.5 \ m

   The diameter of the sprocket is  d_c  =  8.50 \ cm =0.085 \ m

  The angular acceleration considered is  \alpha  =  4.30\ rad/s^2

Generally the radius of the wheel is

       r = \frac{d}{2}

=>     r = \frac{0.5}{2}

=>     r = 0.25 \ m

Generally the radius of the sprocket is

       r_c = \frac{d_c}{2}

=>     r_c = \frac{0.085}{2}

=>     r_c = 0.0425 \ m

Generally the moment of inertia of the wheel is mathematically represented as

      I  =  m  *  r^2

=>    I  =  1.80  *  0.25^2

=>    I  = 1.1125 \ kg \cdot m^2

Generally the torque experienced by the wheel due to the forces acting on it  is mathematically represented as

      \tau =  F_c *  r_c  -  F_t  * r

Here  F_c is the force acting on the sprocket

So  

      \tau =  F_c *  0.0425 - 115  * 0.25

       \tau = 0.0425F_c  -  28.75

Generally the torques that will cause the wheel to move with \alpha  =  4.30\ rad/s^2 is mathematically represented as

       \tau  =  I  * \alpha

So

        0.0425F_c  -  28.75  =   I  * \alpha

        0.0425F_c  -  28.75  =   1.1125  *4.30    

       0.0425F_c  -  28.75  =   1.1125  *4.30    

        F_c  =  789.03 \  N    

5 0
3 years ago
WORTH 47 POINTS!!! It's EASY too!
Vikentia [17]
The Factors are Temperature and kinetic energy, the temperature is because the particles are going to move fast which means the particles in a solid container or in solid pattern they will actually vibrate and they will expand , when temperature increase, more kinetic energy between the particles .
examples is that in the steel when you heat a steel the particles inside it will vibrate then the particles speed up because the vibration increases, therefore the temperature increases so a a thermal expansion occurs that the vibration of the particles will take up more space so the steel bar expands slightly in all Direction if the temperature Falls the reverse happens and the material or steel contracts which means get smaller .
another example is the thermometer, the thermometer has a liquid inside it which is Mercury or alcohol this liquid expands when the temperature rises, the tube is made narrow so that a small increase in volume of the liquid produces a large movement along the tube.

5 0
3 years ago
A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate
vivado [14]

Answer: The molar heat capacity of aluminum is 25.3J/mol^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of water = 130.0 g

m_2 = mass of aluminiunm = 23.5 g

T_{final} = final temperature = 26.0^oC=(273+26)K=299K

T_1 = temperature of water = 23^oC=(273+23)K=296K

T_2 = temperature of aluminium = 100^oC=273+100=373K

c_1 = specific heat of water= 4.184J/g^0C

c_2 = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]

c_2=0.938J/g^0C

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity =0.938J/g^0C\times 27g/mol=25.3J/mol^0C

5 0
3 years ago
The work that a force does by acting on an object is equal to what?
maw [93]
2nd and only 2nd option is right
3 0
3 years ago
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